Extracting Square Roots - Examples, Exercises and Solutions

We first rearrange and then set the equations to equal zero.

$x^2-49=0$

$x^2-7^2=0$

We use the abbreviated multiplication formula:

$(x-7)(x+7)=0$

$x=\pm7$

First, we move the terms to one side equal to 0.

$2x^2-x^2-8-4=0$

We simplify :

$x^2-12=0$

We use the shortcut multiplication formula to solve:

$x^2-(\sqrt{12})^2=0$

$(x-\sqrt{12})(x+\sqrt{12})=0$

$x=\pm\sqrt{12}$

Please note that the equation can be arranged differently:

x²-16 = x +4

x² - 4² = x +4

We will first factor a trinomial for the section on the left

(x-4)(x+4) = x+4

We will then divide everything by x+4

(x-4)(x+4) / x+4 = x+4 / x+4

x-4 = 1

x = 5

The equation in the problem is:

$x^2-x=0$

First let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x$and this is because the first power is the lowest power in the equation and therefore is included both in the term with the second power and in the term with the first power, any power higher than this is not included in the term with the first power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^2-x=0 \\ \downarrow\\ x(x-1)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$\boxed{x=0}$

or:

$x-1=0\\ \downarrow\\ \boxed{x=1}$

Let's summarize then the solution to the equation:

$x^2-x=0 \\ \downarrow\\ x(x-1)=0 \\ \downarrow\\ x=0 \rightarrow\boxed{ x=0}\\ x-1=0 \rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore the correct answer is answer B.

Let's solve the given equation:

$x^2-81=0$Note that we can factor the expression on the left side using the difference of squares formula:

$(\textcolor{red}{a}+\textcolor{blue}{b}) (\textcolor{red}{a}-\textcolor{blue}{b})=\textcolor{red}{a}^2-\textcolor{blue}{b}^2$We'll do this using the fact that:

$81=9^2$Therefore, we'll represent the rightmost term as a squared term:

$x^2-81=0 \\ \downarrow\\ \textcolor{red}{x}^2-\textcolor{blue}{9}^2=0$If so, we can represent the expression on the left side in the above equation as a product of expressions:

$\textcolor{red}{x}^2-\textcolor{blue}{9}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{9})(\textcolor{red}{x}-\textcolor{blue}{9})=0$From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore, we'll get two simple equations and solve them by isolating the variable in each:

$x+9=0\\ \boxed{x=-9}$or:

$x-9=0\\ \boxed{x=9}$

Let's summarize the solution of the equation:

$x^2-81=0 \\ \downarrow\\ \textcolor{red}{x}^2-\textcolor{blue}{9}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{9})(\textcolor{red}{x}-\textcolor{blue}{9})=0 \\ x+9=0\rightarrow\boxed{x=-9}\\ x-9=0\rightarrow\boxed{x=9}\\ \downarrow\\ \boxed{x=9,-9}$Therefore, the correct answer is answer B.

The equation in the problem is:

$3x^2+9x=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $3x$because the first power is the lowest power in the equation and therefore is included both in the term with the second power and in the term with the first power. Any power higher than this is not included in the term with the first power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the letters,

For the numbers, note that 9 is a multiple of 3, therefore it is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$3x^2+9x=0 \\ \downarrow\\ 3x(x+3)=0$

Let's continue and address the fact that on the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$3x=0 \hspace{8pt}\text{/}:3\\ \boxed{x=0}$

In solving the above equation, we divided both sides of the equation by the term with the variable,

Or:

$x+3=0 \\ \boxed{x=-3}$

Let's summarize the solution of the equation:

$3x^2+9x=0 \\ \downarrow\\ 3x(x+3)=0 \\ \downarrow\\ 3x=0 \rightarrow\boxed{ x=0}\\ x+3=0\rightarrow \boxed{x=-3}\\ \downarrow\\ \boxed{x=0,-3}$

Therefore the correct answer is answer C.

The equation in the problem is:

$x^5-4x^4=0$

First note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $x^4$since the fourth power is the lowest power in the equation and therefore is included both in the term with the fifth power and in the term with the fourth power, any power higher than this is not included in the term with the fourth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^5-4x^4=0 \\ \downarrow\\ x^4(x-4)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^4=0 \hspace{8pt}\text{/}\sqrt[4]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$(In this case taking the even root of the right side of the equation will yield two possibilities - positive and negative, however since we're dealing with zero, we get only one solution)

Or:

$x-4=0\\ \downarrow\\ \boxed{x=4}$

Let's summarize the solution of the equation:

$x^5-4x^4=0 \\ \downarrow\\ x^4(x-4)=0 \\ \downarrow\\ x^4=0 \rightarrow\boxed{ x=0}\\ x-4=0 \rightarrow \boxed{x=4}\\ \downarrow\\ \boxed{x=0,4}$

Therefore the correct answer is answer C.

The equation in the problem is:

$x^6+x^5=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $x^5$because the fifth power is the lowest power in the equation and therefore is included both in the term with the sixth power and in the term with the fifth power. Any power higher than this is not included in the term with the fifth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression. We will continue and perform the factoring:

$x^6+x^5=0 \\ \downarrow\\ x^5(x+1)=0$

Let's continue and address the fact that on the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^5=0 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \boxed{x=0}$(in this case taking the odd root of the right side of the equation will yield one possibility)

or:

$x+1=0\\ \boxed{x=-1}$

Let's summarize the solution of the equation:

$x^6+x^5=0 \\ \downarrow\\ x^5(x+1)=0 \\ \downarrow\\ x^5=0 \rightarrow\boxed{ x=0}\\ x+1=0 \rightarrow \boxed{x=-1}\\ \downarrow\\ \boxed{x=0,-1}$

Therefore the correct answer is answer A.

The equation in the problem is:

$x^7-x^6=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x^6$since the sixth power is the lowest power in the equation and therefore is included both in the term with the seventh power and in the term with the sixth power. Any power higher than this is not included in the term with the sixth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^7-x^6=0 \\ \downarrow\\ x^6(x-1)=0$

Let's continue and address the fact that in the left side of the equation we got from the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^6=0 \hspace{8pt}\text{/}\sqrt[6]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$

(in this case taking the even root of the right side of the equation will yield two possibilities - positive and negative but since we're dealing with zero, we get only one answer)

or:

$x-1=0\\ \downarrow\\ \boxed{x=1}$

Let's summarize the solution of the equation:

$x^7-x^6=0 \\ \downarrow\\ x^6(x-1)=0 \\ \downarrow\\ x^6=0 \rightarrow\boxed{ x=0}\\ x-1=0 \rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore the correct answer is answer C.

The equation in the problem is:

$4x^4-12x^3=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $4x^3$because the third power is the lowest power in the equation and therefore is included in both the term with the fourth power and the term with the third power. Any power higher than this is not included in the term with the third power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor for the variables,

For the numbers, note that 12 is a multiple of 4, therefore 4 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$4x^4-12x^3=0 \\ \downarrow\\ 4x^3(x-3)=0$

Let's continue and address the fact that in the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$4x^3=0 \hspace{8pt}\text{/}:4\\ x^3=0 \hspace{8pt}\text{/}\sqrt[3]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the unknown and then extracted a cube root for both sides of the equation.

(In this case, extracting an odd root for the right side of the equation yielded one possibility)

Or:

$x-3=0\\ \boxed{x=3}$

Let's summarize the solution of the equation:

$4x^4-12x^3=0 \\ \downarrow\\ 4x^3(x-3)=0 \\ \downarrow\\ 4x^3=0 \rightarrow\boxed{ x=0}\\ x-3=0\rightarrow \boxed{x=3}\\ \downarrow\\ \boxed{x=0,3}$

Therefore the correct answer is answer A.

The equation in the problem is:

$7x^{10}-14x^9=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and variables in this case is $7x^9$since the ninth power is the lowest power in the equation and therefore is included in both the term with the tenth power and the term with the ninth power. Any power higher than this is not included in the term with the ninth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 14 is a multiple of 7, therefore 7 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0$

Let's continue and address the fact that on the left side of the equation we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get a result of 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$7x^9=0 \hspace{8pt}\text{/}:7\\ x^9=0 \hspace{8pt}\text{/}\sqrt[9]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the variable, and then we extracted a ninth root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

$x-2=0 \\ \boxed{x=2}$

Let's summarize the solution of the equation:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0\\ \downarrow\\ 7x^9=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore, the correct answer is answer A.

The equation in the problem is:

$7x^3-x^2=0$

First, let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x^2$since the second power is the lowest power in the equation and therefore is included both in the term with the third power and in the term with the second power. Any power higher than this is not included in the term with the second power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^2=0 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$(in this case taking the even root of the right side of the equation will indeed yield two possibilities, positive and negative, but since we're dealing with zero, we'll get only one possibility)

or:

$7x-1=0$Let's solve this equation to get the additional solutions (if they exist) to the given equation:

We got a simple first-degree equation which we'll solve by isolating the unknown on one side, we'll do this by moving terms and then dividing both sides of the equation by the coefficient of the unknown:

$7x-1=0 \\ 7x=1\hspace{8pt}\text{/}:7\\ \boxed{x=\frac{1}{7}}$

Let's summarize the solution of the equation:

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0\\ \downarrow\\ x^2=0 \rightarrow\boxed{ x=0}\\ 7x-1=0\rightarrow \boxed{x=\frac{1}{7}}\\ \downarrow\\ \boxed{x=0,\frac{1}{7}}$

Therefore the correct answer is answer C.

First, we should notice that it is a quadratic equation because there is a quadratic term (meaning raised to the second power).

The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side.

Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.

The equation in the problem is already arranged, so let's proceed with the solving technique:

We'll choose to solve it using the quadratic formula.

Let's recall it first:

The rule states that the roots of an equation in the form$ax^2+bx+c=0$ are $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

$$This formula is called: "The Quadratic Formula"

Let's now solve the problem:

$4x^2+1=0$

First, let's identify the coefficients of the terms:

$\begin{cases}a=4 \\ b=0 \\ c=1\end{cases}$

Note that in the given equation there is no first-power term, so from comparing to the general form:

$ax^2+bx+c=0$

we can conclude that the coefficient $b$ (which is the coefficient of the first-power term $x$ in the general form) is 0.

Let's continue and get the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{0\pm\sqrt{0^2-4\cdot4\cdot1}}{2\cdot4}$

Let's continue and calculate the expression under the root and simplify the expression:

$x_{1,2}=\frac{\pm\sqrt{-16}}{8}$

We now have a negative expression under the root and since we cannot extract a real root from a negative number, this equation has no real solutions.

In other words, there is no real value of $x$ that when substituted in the equation will give a true statement.

Therefore, the correct answer is answer D.

The equation in the problem is:

$x^{100}-9x^{99}=0$

First, note that in the left side we can factor out a common factor from the terms, the largest common factor for the numbers and letters in this case is $x^{99}$because the power of 99 is the lowest power in the equation and therefore is included both in the term with power of 100 and in the term with power of 99. Any power higher than this is not included in the term with the lowest power of 99, and therefore this is the term with the highest power that can be factored out as a common factor from all letter terms,

Let's continue and perform the factoring:

$x^{100}-9x^{99}=0 \\ \downarrow\\ x^{99}(x-9)=0$

Let's continue and address the fact that on the left side of the equation we received in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get a result of 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$x^{99}=0 \hspace{8pt}\text{/}\sqrt[99]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we extracted a 99th root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

$x-9=0 \\ \boxed{x=9}$

Let's summarize the solution of the equation:

$x^{100}-9x^{99}=0 \\ \downarrow\\ x^{99}(x-9)=0 \\ \downarrow\\ x^{99}=0 \rightarrow\boxed{ x=0}\\ x-9=0\rightarrow \boxed{x=9}\\ \downarrow\\ \boxed{x=0,9}$

Therefore, the correct answer is answer C.

The equation in the problem is:

$x^7-5x^6=0$

First let's note that in the left side we can factor the expression using a common factor, the largest common factor for the numbers and letters in this case is $x^6$since the sixth power is the lowest power in the equation and therefore is included both in the term with the seventh power and in the term with the sixth power, any power higher than this is not included in the term with the sixth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression, so we'll continue and perform the factoring:

$x^7-5x^6=0 \\ \downarrow\\ x^6(x-5)=0$

Let's continue and address the fact that in the left side of the equation we got in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to get 0 from multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^6=0 \hspace{8pt}\text{/}\sqrt[6]{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$

(in this case taking the even root of the right side of the equation will yield two possibilities - positive and negative, however since we're dealing with zero, we get only one solution)

or:

$x-5=0\\ \downarrow\\ \boxed{x=5}$

Let's summarize the solution of the equation:

$x^7-5x^6=0 \\ \downarrow\\ x^6(x-5)=0 \\ \downarrow\\ x^6=0 \rightarrow\boxed{ x=0}\\ x-5=0 \rightarrow \boxed{x=5}\\ \downarrow\\ \boxed{x=0,5}$

Therefore the correct answer is answer A.